Theorem: Let $\{C_1, \ldots, C_m\}$ be convex sets in $\mathbb{R}^n$ with $m \geq n+1$. If every $(n+1)$ of them intersect, then all $m$ sets intersect.

Proof (by induction on $m$):

Base case ($m = n+1$): This is the hypothesis itself.

Inductive step: Assume the theorem holds for $m-1$ sets, and consider $m$ sets $C_1, \ldots, C_m$ where every $(n+1)$ intersect.

Step 1: For $i = 1, \ldots, m$, define:

By hypothesis, every $n$ of the sets $D_1, \ldots, D_m$ have non-empty intersection (take $n$ sets $D_i$ and use that the corresponding $(n+1)$ of the original $C_j$ intersect).

Step 2: By induction hypothesis applied to $D_1, \ldots, D_{n+1}$, there exists a point:

Step 3: Point $p$ lies in $D_1 \cap \cdots \cap D_{n+1}$. For each $i \in \{1, \ldots, n+1\}$:

This means $p$ lies in all $C_j$ except possibly $C_i$. But $p$ must lie in at least $m - (n+1)$ of the sets $C_1, \ldots, C_m$.

Step 4: Actually, $p$ lies in all $C_i$: For any particular $C_k$, consider $D_i$ for $i \neq k$. Since $p \in D_i$ for $i \leq n+1$, and these cover all indices except possibly $k$, we have $p \in C_k$ as well.

Therefore, $p \in \bigcap_{i=1}^m C_i$. ∎